(4x^2+5x-8)+(2x^2-x+7=)

Solution for (4x^2+5x-8)+(2x^2-x+7=) equation:

(4x^2+5x-8)+(2x^2-x+7=)

We move all terms to the left:

(4x^2+5x-8)+(2x^2-x+7-())=0

We get rid of parentheses

4x^2+5x+(2x^2-x+7-())-8=0


We calculate terms in parentheses: +(2x^2-x+7-()), so:

2x^2-x+7-()

We add all the numbers together, and all the variables

2x^2-1x

Back to the equation:

+(2x^2-1x)


We get rid of parentheses

4x^2+2x^2+5x-1x-8=0

We add all the numbers together, and all the variables

6x^2+4x-8=0

a = 6; b = 4; c = -8;
Δ = b2-4ac
Δ = 42-4·6·(-8)
Δ = 208
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{13}}{2*6}=\frac{-4-4\sqrt{13}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{13}}{2*6}=\frac{-4+4\sqrt{13}}{12}
$